A Derivative?

Question

I'm in calculus II but have forgotten alot of Calc I.

How to find the Derivative of...

2894/(1+259*e^(x/400))
?

For some reason I got 1873.865/(e^(x/400) + 67081/(e^(x/400)) + 518)

There might be a missing division sign in the problem someone else wrote below.

Answer 1

I'll start off by saying here is my work in MSPaint, in which the math is much easier to follow than the text below. The link is:

http://img187.imageshack.us/img187/8219/math1jk8.png




However, the text below explains what I did step by step so it's still there.


Rewrite

2894/(1+259*e^(x/400))

as:

2894(1+259*e^(x/400))^(-1)

and 2894 is a constant, so will ignore that part and just find the derivative of

(1+259*e^(x/400))^(-1)

which will be our "u^n" when we use the chain rule. And since the derivative of u^n = n u^(n-1) u'

where n = -1, u = 1+259*e^(x/400), and u' = 259/400 e^(x/400)

Do you understand how I found u' ? Keep in mind d/dx of e^u = (e^u)(u') and in this part of the problem, u = x/400, u' = 1/400

and the derivative of a constant, 1, is zero, so that part before the plus sign goes away, and we multiply by 259 since the constant is being multiplied by the derivative which gives us 259/400 e^(x/400)

Okay so using our chain rule:

n = -1, u = 1+259*e^(x/400), and u' = 259/400 e^(x/400)
u^n = n u^(n-1) u'

(1+259*e^(x/400))^(-1) = (-1)[(1+259*e^(x/400))^(-2)][259/400 e^(x/400)]

= -259/400 ((1+259*e^(x/400))^(-2)) e^(x/400)

And don't forget to multiply by the constant, 2894, that we factored out in the very beginning of the problem...

= -(259*2894)/400 ((1+259*e^(x/400))^(-2)) e^(x/400)

Or in another form:

= [374773 e^(x/400)] / [200(1+259e^(x/400)]^(-2)

Again, I also did this in MSPaint if you have a hard time following math symbols in text:
http://img187.imageshack.us/img187/8219/math1jk8.png

Best of luck to you. Chain Rule is so much easier than product/quotient rule anyway.

Answer 2

It's an example of using the chain-rule, i.e. that

d/dx f(g(h(x))) = [df(g(h(x)))/dg(h(x))] [dg(h(x))/dh(x)] [dh(x)/dx], etc.

So, using this approach,

d/dx [ 2894/(1+259*e^(x/400)) ]

= 2894 [- 1 / (1+259*e^(x/400))^2] [259*e^(x/400)] (1 / 400)

= - (259*2894 / 400) e^(x/400) / (1+259*e^(x/400))^2.

QED

Live long and prosper.

Answer 3

let f(x)= 2894/ ( 1 + 259* e^x/400)
( u/v)'= vu'-uv'/v^2

derfore,
f'(x) = -2894( 0 + 259* e^x/400 .1/400 )/ (denominator)^2. ( using chain rule, derivative of e^x/400 = e^x/400 * (derivative of x/400)...
hope dis helped

Answer 4

d f(x)/g(x) /dx = [ g(x) f'(x) - g'(x) f(x) ] / [g(x)]^2

f(x) = 2894
f'(x) = 0

g(x) = 1 + 259*e^(x/400)
g'(x) = 259/400 * e^(x/400)

I'm going to be lazy and leave the rest of the mess for you to solve.

Answer 5

http://mathforum.org/library/drmath/sets/select/dm_derivative.htm

ask dr. math. I took calc a long time ago too. sorry. it is very difficult to explain in text, it won't be fun to read. Only other suggestion is to go to the math clinic at the college.

Answer 6

no need to find...
the question is not going to come in the exams...
so dont worry....